00:01
Hi, in this question we have a steel cylindrical plug given with a figure where t is the torque applied.
00:08
So, now we have the expression for dt is given as it is pr d theta lr cos theta minus pr d theta lr sin theta plus mu into pr d theta lr.
00:24
Now integrating this expression from 0 to 2 pi dt is equal to it is integration 0 to 2 pi pr square l cos theta into d theta minus integration pr square l sin theta d theta and plus integration mu pr square into l d theta.
00:55
Now, so integration of dt is t which is equal to pr square l.
01:02
So, integration of cos theta is sin theta from 0 to 2 pi minus pr square l.
01:10
So, integration of sin theta is minus cos theta minus minus plus.
01:15
So, this is cos theta from 0 to 2 pi plus integration of mu pr square is common l integration of d theta is theta from 0 to 2 pi.
01:27
Now applying the lower limit and the upper limit we have.
01:31
So, this is 0 because integration sin 0 is 0 and sin 2 pi is 0.
01:37
So, here cos 0 is 1 and cos 2 pi is 1.
01:43
So, 1 minus 1 will be equal to 0.
01:45
So, this is the remaining term which is mu pr square l into 2 pi...